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lilavasa [31]
3 years ago
10

Factor the following , it is not prime

Mathematics
1 answer:
MrMuchimi3 years ago
4 0
So 1/9-16z^2 this is a diffirence of two perfect squares thing
so (1/3)^2-(4z)^2
so ((1/3)-4z)((1/3)+4z)

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COMMON CORE REVIEW
masha68 [24]

Answer:

146.9

Step-by-step explanation:

I'm not sure of the answer but if I'm wrong tell me I'll continue trying it is a pleasure to help you.

7 0
3 years ago
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}


                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}


                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}


                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
3 years ago
If it takes 13 cup of flour to make 14 of a cake, how many cups of flour are needed to make a whole cake? i need help plz
aleksandrvk [35]

Answer:

1-1/3 if its 14

1/52 if its 1/4

Step-by-step explanation:

4 0
3 years ago
10 points for however can tell me what coler of horse is this and a brainliest
posledela

Answer:

dark brown and white

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Express sin UU as a fraction in simplest terms.
alukav5142 [94]

The expression of sinU as a fraction is 12/13

Find the diagram attached.

To get the fraction represented by sinU, we will use the SOH CA TOA identity.

From the diagram;

  • Hypotenuse = SU = 13
  • Opposite = ST = 12 (angle opposite to m<U)

Since sin theta = opp/hyp

SinU = ST/SU

SinU = 12/13

Hence the expression of sinU as a fraction is 12/13

Learn  more on SOH CAH TOA here: brainly.com/question/20734777

5 0
2 years ago
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